Monday, March 1, 2010

The fundamental group functor part 2

So... in the previous post I promised to finish what I was saying about the fundamental group functor... So far I've sketched the proof that this is, indeed, a functor. I would show the proof in detail, but it's long, tedious and not very informative-- the point is, it's a map from Top$_*$ to Grp which preserves morphisms. There are much more interesting things a functor can preserve. Namely, it can preserve products and coproducts.

So what's a product? As a motivating example, look at Set. When we talk about the product of two sets, we clearly mean the cartesian product. Since we're interested in category theory at the moment, we don't really want to talk about members of the product, we want to talk about maps to and from the product.

It turns out there are two really nice maps $pr_1:A\times B\rightarrow A$ and $pr_2: A\times B\rightarrow B$, the projections onto $A$ and $B$ respectively. It turns out that they have a really nice universal property: Given an object $V$, and two maps $f:V\rightarrow A$, $g:V\rightarrow B$, we can "factor" $f$ and $g$ through $A\times B$ in a unique way. This means that we have a unique map $h:V\rightarrow A\times B$ such that $f=pr_1\circ h$ and $g=pr_2\circ h$. At first it may be a bit surprising that this map (sometimes called $f\times g$) is unique. But really, our two projections forget everything about one side of our product, so the function needs to act "independently" on $A$ and $B$, and there's really only one way to get this to interact properly with the projections.

Something that's more surprising is this: the product is unique up to unique isomorphism. This means that if there is a "different" product (Why not try $B\times A$?), there is a single, canonical isomorphism between the two objects-- just factor the projections from one product through the other. This map is unique, and it damned-well better be an isomorphism. (To see that it is, factor the projections back the other way, wave your hands about and say something about "the identity morphism".)

Ok. So, by analogy with Set, we (sort of) get what a product is. What about coproducts? A nice thing about category theory is that whenever you see a word that starts with "co", you can figure out what it means in 3 easy steps:

remove the "co" from the word.

Draw the diagram that represents the word you just found.

Turn around all the arrows.

So, this means the coproduct, $A\coprod B$, should have two maps $i_1:A\rightarrow A\coprod B$ and $i_2:B\rightarrow A\coprod B$ (called the imbeddings) such that for any pair of maps $f:A\rightarrow V$ and $g:B\rightarrow V$, we have a unique morphism $h:A\coprod B\rightarrow V$ such that $h\circ i_1 = f$ and $h\circ i_2 = g$. For some reason this always seems a little harder to follow. Let's work it out in Set. Let's look at $f$ and $g$ as in the definition. We want some map (call it $f*g$ because I can't think of what the actual notation is) that goes from somewhere to $V$ such that $f*g \circ i_1 = f$ and the same with $g$. We want $i_1$ and $i_2$ to do almost nothing... what happens if we take $A\coprod B$ as the disjoint union? (Hence the notation...) What is the imbedding? It's the "move me from $A$ to $A\coprod B$" function. And what could $f*g$ possibly be? Well obviously, it's the function which sends $a\in A\mapsto f(a)$ and $b\in B\mapsto g(b)$.

Ok. Great. We know what the product and coproduct need to look like (at least when we only care about the product of two objects). What exactly are they in Top$_*$? It turns out they are the wedge products-- take the disjoint union (familiar?) and glue the two spaces together at their base-points. This means that we have two completely unrelated spaces (modulo open sets containing the basepoint.)

This idea of a coproduct being the result of "smashing together" two objects without making them at all related is basically consistent throughout basically every category. In fact, in the category of groups, it's the free product, which is the "freely generated" product of the two groups-- For groups $G$ and $H$, this means the set of all words on $G\cup H$, where things reduce in the "obvious" way and no other way... I'm going to pretend like this makes sense to you, since (as you've surely learned by now) I have yet to decide what level of audience I'm writing for.

So, taking this back to the fundamental group functor: for two (pointed) spaces $(S,s)$ and $(T,t)$, we would like $\pi_1(S\times T, (s,t)) = \pi_1(S,s)\times \pi_1(T,t)$ and $\pi_1(S\coprod T, (s,t)) = \pi_1(S,s)\coprod \pi_1(T,t)$.

Guess what? I'm going to cop out of actually proving this! (Are you surprised? You should be used to this by now...)

However, I will at least wave my hands around a bit and give you a feel for why it's true. First let's look at products. As an example, look at the torus-- $S^1\times S^1$. Draw a path on this. We want to be able to push this path down to a path which only lives in one copy of $S^1$ in each component. (I.e., a path which stays on $(S^1\times\{0\})\cup(\{0\}\times S^1)$. ) We can do this by pushing (in a continuous fashion-- i.e., homotopically) all points of our path onto one of our two reference circles.

For coproducts: it's a little more obvious in some sense--- any path is going to stay in one of our two spaces for a while, and then cross over to the other. The homotopy group we get here "reduces" in the obvious way, and no way else-- i.e. it's the free product.